A particle is projected vertically upwards and reaches the maximum height $H$ at a time $t = T$ . The height of the particle at any time $t (< T)$ will be

g {(t - T)}^{2}

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H - g {(t - T)}^{2}

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\frac{1}{2} g {(t - T)}^{2}

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H - \frac{1}{2} g {(T - t)}^{2}

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Solution

The correct option is D $H - \frac{1}{2} g {(T - t)}^{2}$
At maximum height $, v = 0$
Thus, from equations of motion:
$0 = u - g T$ or $u = g T$
$H = u T - \frac{1}{2} g T^{2} = g T^{2} - \frac{1}{2} g T^{2}$
or, $H = 0.5 g T^{2}$
At any time $, t, h = u t - 0.5 g t^{2} = g T t - 0.5 g t^{2}$
Add and subtract $0.5 g T^{2}$ , we get,
$h = (g t T - 0.5 g t^{2} - 0.5 g T^{2}) + 0.5 g T^{2}$
Put $g T^{2} = H$ , we get
$h = - 0.5 g (T - t)^{2} + H = H - \frac{1}{2} g (T - t)^{2}$

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A particle is projected vertically upwards and reaches the maximum height H at a time t=T. The height of the particle at any time t(<T) will be

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